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Math Help - sequence

  1. #1
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    sequence

    Prove the following: Let  (a_n) and  (b_n) be sequences of real numbers that converge to real numbers  a and  b respectively. If  a_n \leq b_n for all  n \in \mathbb{N} , then  a \leq b .

    Proof: For each positive real number  \epsilon there exists a  N_1 \in \mathbb{N} so that whenever  n > N_1 ,  |a_n-a| < \epsilon . Also for each positive real number  \epsilon there exists a  N_2 \in \mathbb{N} so that whenever  n > N_2  ,  |b_n-b| < \epsilon . So  a- \epsilon < a < a + \epsilon and  b- \epsilon < b < b+ \epsilon . Then  0 \leq b_n- a_n < b-a for all  n \in \mathbb{N} . It follows that  a \leq b .  \square

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  2. #2
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    Quote Originally Posted by manjohn12 View Post
    Prove the following: Let  (a_n) and  (b_n) be sequences of real numbers that converge to real numbers  a and  b respectively. If  a_n \leq b_n for all  n \in \mathbb{N} , then  a \leq b .

    Proof: For each positive real number  \epsilon there exists a  N_1 \in \mathbb{N} so that whenever  n > N_1 ,  |a_n-a| < \epsilon . Also for each positive real number  \epsilon there exists a  N_2 \in \mathbb{N} so that whenever  n > N_2  ,  |b_n-b| < \epsilon . So  a- \epsilon < a < a + \epsilon and  b- \epsilon < b < b+ \epsilon . Then  0 \leq b_n- a_n < b-a for all  n \in \mathbb{N} . It follows that  a \leq b .  \square

    Is this correct?
    Try proving the following easier result: let \{ x_n\} be a sequence such that x_n \geq 0 and x_n \to x then x\geq 0.

    After you prove that consider x_n = a_n - b_n \geq 0.
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